1
JEE Advanced 2013 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

A right-angled prism of refractive index $$\mu$$1 is placed in a rectangular block of refractive index $$\mu$$2, which is surrounded by a medium of refractive index $$\mu$$3, as shown in the figure. A ray of light e enters the rectangular block at normal incidence. Depending upon the relationships between $$\mu$$1, $$\mu$$2 and $$\mu$$3, it takes one of the four possible paths 'ef', 'eg', 'eh' or 'ei'.

JEE Advanced 2013 Paper 2 Offline Physics - Geometrical Optics Question 30 English

Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists:

List I List II
P. $$e \to f$$
1. $${\mu _1} > \sqrt 2 {\mu _2}$$
Q. $$e \to g$$
2. $${\mu _2} > {\mu _1}$$ and $${\mu _2} > {\mu _3}$$
R. $$e \to h$$
3. $${\mu _1} = {\mu _2}$$
S. $$e \to i$$
4. $${\mu _2} < {\mu _1} < \sqrt 2 {\mu _2}$$ and $${\mu _2} > {\mu _3}$$

A
P-2, Q-3, R-1, S-4
B
P-1, Q-2, R-4, S-3
C
P-4, Q-1, R-2, S-3
D
P-2, Q-3, R-4, S-1
2
JEE Advanced 2013 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-0.75
The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $${2 \over 3}$$ times the wavelength in free space. The radius of the curved surface of the lens is
A
1 m
B
2 m
C
3 m
D
6 m
3
JEE Advanced 2013 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

A ray of light travelling in the direction $${1 \over 2}\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ is incident on a plane mirror. After reflection, it travels along the direction $${1 \over 2}\left( {\widehat i - \sqrt 3 \widehat j} \right)$$. The angle of incidence is

A
30$$^\circ$$
B
45$$^\circ$$
C
60$$^\circ$$
D
75$$^\circ$$
4
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, $${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation $$n = \left( {{c \over v}} \right) = \pm \,\sqrt {{\varepsilon _r}{\mu _r}} $$, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $$\varepsilon $$r and $$\mu$$r, are the relative permittivity and permeability of the medium, respectively.

In normal materials, both $$\varepsilon$$r and $$\mu$$r , are positive, implying positive n for the medium. When both $$\varepsilon$$r and $$\mu$$r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating and physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

For light incident from air on a meta-material, the appropriate ray diagram is

A
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 27 English Option 1
B
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 27 English Option 2
C
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 27 English Option 3
D
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 27 English Option 4
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